s=Σ_(i=1) ^∞ 2^i 2s=s−1 s=1 ? how to explain it and how to judge which case can use this way


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Question Number 210788 by liuxinnan last updated on 19/Aug/24

$s = \underset{i = 1}{\sum^{\infty}} 2^{i}$ $2 s = s - 1$ $s = 1 ?$ $h o w t o e x p l a i n i t$ $a n d h o w t o j u d g e w h i c h c a s e c a n u s e t h i s w a y$
	Answered by Berbere last updated on 19/Aug/24

	$w h e n Y o u h a v e e q a t i o n$ $t h e v a r i a b l e i s i n s e t$ $2 s = s - 1 h a v e s \in R . . ?$ $R =] - \infty, \infty [$ $s = \underset{i = 1}{\sum^{\infty}} 2^{i} > 2^{n}; \forall n \in N$ $s \to \infty \notin R$
	Commented by liuxinnan last updated on 20/Aug/24

	$t h a n k s$
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